Question

If $a,b,c,x,y,z\in \mathbb{R}$ then

{{\left( a-x \right)}^{2}} & {{\left( b-x \right)}^{2}} & {{\left( c-x \right)}^{2}} \\\ {{\left( a-y \right)}^{2}} & {{\left( b-y \right)}^{2}} & {{\left( c-y \right)}^{2}} \\\ {{\left( a-z \right)}^{2}} & {{\left( b-z \right)}^{2}} & {{\left( c-z \right)}^{2}} \\\ \end{matrix} \right|=\left| \begin{matrix} {{\left( 1+ax \right)}^{2}} & {{\left( 1+bx \right)}^{2}} & {{\left( 1+cx \right)}^{2}} \\\ {{\left( 1+ay \right)}^{2}} & {{\left( 1+by \right)}^{2}} & {{\left( 1+cy \right)}^{2}} \\\ {{\left( 1+az \right)}^{2}} & {{\left( 1+bz \right)}^{2}} & {{\left( 1+cz \right)}^{2}} \\\ \end{matrix} \right|$$ (a) True (b) False

Answer 1

We solve this problem by using the product of determinants. We take the LHS and expand the square and we divide it into the product of two determinants.
We use the formulas of square of sum and difference of two numbers that is
${{\left( x+y \right)}^{2}}={{x}^{2}}+2xy+{{y}^{2}}$
${{\left( x-y \right)}^{2}}={{x}^{2}}-2xy+{{y}^{2}}$
Then we use the possible transformations of rows and columns to get the RHS. We use one of the standard conditions that when we interchange the columns or rows in odd number of times then we get a negative sign out.

Complete step by step answer:
We are given that

{{\left( a-x \right)}^{2}} & {{\left( b-x \right)}^{2}} & {{\left( c-x \right)}^{2}} \\\ {{\left( a-y \right)}^{2}} & {{\left( b-y \right)}^{2}} & {{\left( c-y \right)}^{2}} \\\ {{\left( a-z \right)}^{2}} & {{\left( b-z \right)}^{2}} & {{\left( c-z \right)}^{2}} \\\ \end{matrix} \right|=\left| \begin{matrix} {{\left( 1+ax \right)}^{2}} & {{\left( 1+bx \right)}^{2}} & {{\left( 1+cx \right)}^{2}} \\\ {{\left( 1+ay \right)}^{2}} & {{\left( 1+by \right)}^{2}} & {{\left( 1+cy \right)}^{2}} \\\ {{\left( 1+az \right)}^{2}} & {{\left( 1+bz \right)}^{2}} & {{\left( 1+cz \right)}^{2}} \\\ \end{matrix} \right|$$ Let us assume that the LHS as $$\Rightarrow \Delta =\left| \begin{matrix} {{\left( a-x \right)}^{2}} & {{\left( b-x \right)}^{2}} & {{\left( c-x \right)}^{2}} \\\ {{\left( a-y \right)}^{2}} & {{\left( b-y \right)}^{2}} & {{\left( c-y \right)}^{2}} \\\ {{\left( a-z \right)}^{2}} & {{\left( b-z \right)}^{2}} & {{\left( c-z \right)}^{2}} \\\ \end{matrix} \right|$$ We know that the formula of square of difference of two numbers as $${{\left( x-y \right)}^{2}}={{x}^{2}}-2xy+{{y}^{2}}$$ By using this formula let us convert the above equation then we get $$\Rightarrow \Delta =\left| \begin{matrix} {{a}^{2}}-2ax+{{x}^{2}} & {{b}^{2}}-2bx+{{x}^{2}} & {{c}^{2}}-2cx+{{x}^{2}} \\\ {{a}^{2}}-2ay+{{y}^{2}} & {{b}^{2}}-2by+{{y}^{2}} & {{c}^{2}}-2cy+{{y}^{2}} \\\ {{a}^{2}}-2az+{{z}^{2}} & {{b}^{2}}-2bz+{{z}^{2}} & {{c}^{2}}-2cz+{{z}^{2}} \\\ \end{matrix} \right|$$ Now, let us divide the above determinant as the product of two determinants then we get $$\Rightarrow \Delta =\left| \begin{matrix} 1 & x & {{x}^{2}} \\\ 1 & y & {{y}^{2}} \\\ 1 & z & {{z}^{2}} \\\ \end{matrix} \right|\times \left| \begin{matrix} {{a}^{2}} & {{b}^{2}} & {{c}^{2}} \\\ -2a & -2b & -2c \\\ 1 & 1 & 1 \\\ \end{matrix} \right|$$ Now, let us take the negative sign out from the second column of the second determinant then we get $$\Rightarrow \Delta =-\left| \begin{matrix} 1 & x & {{x}^{2}} \\\ 1 & y & {{y}^{2}} \\\ 1 & z & {{z}^{2}} \\\ \end{matrix} \right|\times \left| \begin{matrix} {{a}^{2}} & {{b}^{2}} & {{c}^{2}} \\\ 2a & 2b & 2c \\\ 1 & 1 & 1 \\\ \end{matrix} \right|$$ We know that when we interchange the columns or rows an odd number of times then we get a negative sign out. Now, let us interchange the first and third rows of the second determinant then by using the above condition we get that $$\begin{aligned} & \Rightarrow \Delta =-\left| \begin{matrix} 1 & x & {{x}^{2}} \\\ 1 & y & {{y}^{2}} \\\ 1 & z & {{z}^{2}} \\\ \end{matrix} \right|\times \left( -\left| \begin{matrix} 1 & 1 & 1 \\\ 2a & 2b & 2c \\\ {{a}^{2}} & {{b}^{2}} & {{c}^{2}} \\\ \end{matrix} \right| \right) \\\ & \Rightarrow \Delta =\left| \begin{matrix} 1 & x & {{x}^{2}} \\\ 1 & y & {{y}^{2}} \\\ 1 & z & {{z}^{2}} \\\ \end{matrix} \right|\times \left| \begin{matrix} 1 & 1 & 1 \\\ 2a & 2b & 2c \\\ {{a}^{2}} & {{b}^{2}} & {{c}^{2}} \\\ \end{matrix} \right| \\\ \end{aligned}$$ Now, by multiplying the determinants above we get the combined determinant as $$\Rightarrow \Delta =\left| \begin{matrix} 1+2ax+{{\left( ax \right)}^{2}} & 1+2bx+{{\left( bx \right)}^{2}} & 1+2cx+{{\left( cx \right)}^{2}} \\\ 1+2ay+{{\left( ay \right)}^{2}} & 1+2by+{{\left( by \right)}^{2}} & 1+2cy+{{\left( cy \right)}^{2}} \\\ 1+2az+{{\left( az \right)}^{2}} & 1+2bz+{{\left( bz \right)}^{2}} & 1+2cz+{{\left( cz \right)}^{2}} \\\ \end{matrix} \right|$$ We know that the formula of square of sum of two numbers as $${{\left( x+y \right)}^{2}}={{x}^{2}}+2xy+{{y}^{2}}$$ By using this formula to above determinant we get $$\begin{aligned} & \Rightarrow \Delta =\left| \begin{matrix} {{\left( 1+ax \right)}^{2}} & {{\left( 1+bx \right)}^{2}} & {{\left( 1+cx \right)}^{2}} \\\ {{\left( 1+ay \right)}^{2}} & {{\left( 1+by \right)}^{2}} & {{\left( 1+cy \right)}^{2}} \\\ {{\left( 1+az \right)}^{2}} & {{\left( 1+bz \right)}^{2}} & {{\left( 1+cz \right)}^{2}} \\\ \end{matrix} \right| \\\ & \Rightarrow \Delta =RHS \\\ \end{aligned}$$ Here, we can see that both LHS and RHS are equal. Therefore, we can say that the given equation is true. **So, the correct answer is “Option a”.** **Note:** In this problem the main part is that we need to divide the determinant into the product of two determinants. Here, it is easy that we can get an idea of how to divide the determinants. In both the determinants we get one column in each determinant equal to 1 because we need to place the squares of numbers in one column to get the product as the square of difference of two numbers. So, we get the determinants as $$\Rightarrow \Delta =\left| \begin{matrix} 1 & x & {{x}^{2}} \\\ 1 & y & {{y}^{2}} \\\ 1 & z & {{z}^{2}} \\\ \end{matrix} \right|\times \left| \begin{matrix} {{a}^{2}} & {{b}^{2}} & {{c}^{2}} \\\ -2a & -2b & -2c \\\ 1 & 1 & 1 \\\ \end{matrix} \right|$$ Here, we need to keep in mind that the column in each determinant whose entries will be 1 should be different columns in both determinants . Here in the first determinant we get the third column and in the second determinant we get the first column that has entries as 1.